Question

Find general solution of differential solution given below:

$(1+y^2)dx=({\tan }^{-1}y-x)dy$

• A. $x=c{e}^{arc\cot y}+arc\cot y-1$
• B. $x=c{e}^{arc\tan y}+arc\tan y+1$
• C. $x=c{e}^{-arc\cot y}+arc\cot y+1$
• D. $x=c{e}^{-arc\tan y}+arc\tan y-1$

Answer 1

The correct option is D $x=c{e}^{-arc\tan y}+arc\tan y-1$

$(1+y^2)dx=({\tan }^{-1}y-x)dy$
Substitute $t={\tan }^{-1}y\Rightarrow dt=\frac{1}{1+y^2}dy$
$(1+y^2)dx=(t-x)(1=y^2)dt$
$\Rightarrow \frac{dx}{dt}+x=t$
I.F $=e^{\int 1dt}=e^t$
Thus solution is, $x.e^t=\int t{e}^{t}dt+c=t{e}^t-e^t+c$
$\Rightarrow x=c{e}^{-arc\tan y}+arc\tan y-1$